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13.2.1. Find $\bar{Y}$ and $\sigma_{y}$ for the sample
| $y_{j}$ | 0.5 | 1.3 | 2.1 | 2.9 | 3.7 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $m_{y_{j}}$ | 5 | 20 | 45 | 27 | 3 | | Solution. To calculate $\bar{Y}, D_{y}$, and $\sigma_{y}$, we will use the method of products. Introduce conditional variants $v_{j}=\left(y_{j}-c_{y}\right) / h_{y}$, where $c_{y}$ is the value of $y_{j}$ corresponding to the highest frequency, $c_{y}=2.1$, and $h_{y}$ is the step of the sample, $h_{y}=0.8$.
Then, ca... | \bar{Y}=2.12,\sigma_{y}=0.71 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,429 |
### 7.1.1. Optimal Production Task
For the production of two types of products I and II, three types of raw materials are used.
To produce one unit of product I, it is required to spend 13 kg of the first type of raw material, 32 kg of the second type of raw material, and 58 kg of the third type of raw material.
To ... | Solution. Let's consider the mathematical model of the problem. If we take $x_{1}$ as the number of items I planned for production and $x_{2}$ as the number of items II, we obtain a linear programming problem.
Subject to the constraints
$$
\begin{gathered}
13 x_{1}+24 x_{2} \leq 312 \\
32 x_{1}+32 x_{2} \leq 480 \\
5... | 54 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,430 |
5. Find the lengths of the segments determined on the axis by points $A(2)$ and $B(-5), C(-2)$ and $D(4)$. The first point denotes the beginning of the segment, and the second - its end. | Solution. Applying formula (1), we get:
$$
A B=-5-2=-7 ; C D=4-(-2)=6
$$ | -7,6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,432 |
6. Find the distance $d$ between points $A$ and $B$ in each of the following cases:
1) $A(2), B(3) ; 2) A(-4), B(-8)$. | Solution. Applying formula (2), we get:
1) $d=|3-2|=1 ; 2) d=|-8-(-4)|=4$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,433 |
7. Find the point $C$ that divides the segment between points $A(-1)$ and $B(5)$ in the ratio:
1) $\lambda=\frac{2}{3}$; 2) $\lambda=-2$. | Solution. Applying formula (3), we find the coordinate of point $C$:
1) $x=\frac{-1+\frac{2}{3} \cdot 5}{1+\frac{2}{3}}=1.4$
2) $x=\frac{-1+(-2) \cdot 5}{1+(-2)}=11$.
Thus: 1) $C(1.4)$; 2) $C(11)$. | 1)C(1.4);2)C(11) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,434 |
16. Construct the point $M(-4 ; 3)$. | Solution. On the $O x$ axis, we lay off the segment $\overline{O P}$, the magnitude of which is -4, and on the $O y$ axis - the segment $\overline{O Q}$, which is equal to 3 in magnitude.
We draw a line through point $P$ parallel to the $O y$ axis, and through point $Q$ - a line parallel to the $O x$ axis. The point o... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,435 |
25. Find the distance between the points: $A(2; -3)$ and $B(-1; 1)$; $C(-3; 4)$ and $D(5; -2)$; $O(0; 0)$ and $E(1; -4)$. | Solution. Using formulas (1) and (2), we get: the length of $AB=\sqrt{(-1-2)^{2}+[1-(-3)]^{2}}=\sqrt{25}=5$; the length of $CD=\sqrt{[5-(-3)]^{2}+(-2-4)^{2}}=\sqrt{100}=10$; the length of $OE=\sqrt{1^{2}+(-4)^{2}}=\sqrt{1+16}=\sqrt{17}$. | 5,10,\sqrt{17} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,436 |
26. Find out if there is an obtuse angle among the interior angles of the triangle with vertices $A(1 ; 2), B(-3 ; 4), C(0 ;-2)$. | Solution. Using formula (1), we find the lengths of the sides of the triangle:
$$
\begin{aligned}
& AB=\sqrt{(-3-1)^{2}+(4-2)^{2}}=\sqrt{16+4}=2 \sqrt{5} \\
& BC=\sqrt{3^{2}+(-2-4)^{2}}=\sqrt{9+36}=3 \sqrt{5} \\
& AC=\sqrt{1^{2}+[2-(-2)]^{2}}=\sqrt{1+16}=\sqrt{17}
\end{aligned}
$$
We find the squares of the lengths o... | \angleBAC | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,437 |
39. A line segment with endpoints at points $A(3 ; 2)$ and $B(12 ; 8)$ is divided into three equal parts. Determine the coordinates of the division points and the length of segment $\overline{A B}$. | Solution. We will use formula (1).
For point $C$ (Fig. 7) the
Fig. 7 ratio $\frac{A C}{C B}=\lambda=\frac{1}{2}$. Therefore, the coordinates of point $C$ are expressed as follows:
$$
x=\... | C(6;4),D(9;6),AB=\sqrt{117}\approx10.82 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,438 |
40. Segment $\overline{A B}$ with endpoints at points $A(-3 ; 2)$ and $B(4 ;-5)$ is divided by point $C$ in the ratio $\lambda=-3$. Find the coordinates of the point of division.
Fig. 8 | Solution. Using formula (1), we get:
$$
\begin{aligned}
& x=\frac{-3+(-3) \cdot 4}{1+(-3)}=7.5 \\
& y=\frac{2+(-3)(-5)}{1+(-3)}=-8.5
\end{aligned}
$$
Thus, the desired point $C(7.5$; $-8.5$) divides the segment $\overline{A B}$ externally in the ratio $\frac{A C}{C B}=$ $=\lambda=-3$ (Fig. 8). Note that the lengths $... | C(7.5,-8.5) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,439 |
58. Find the equation of the circle with center at point $C(2 ; 3)$ and radius $R=5$ (Fig. 9). Do the points $A(5 ; 7), B(6 ; 0), E(-1 ; 5)$ and $D(1 ;-3)$ lie on this circle? | Solution. By definition of a circle, the distance from any point $M(x ; y)$ lying on it to its center $C(2 ; 3)$ is equal to the length of the radius $R=5$, i.e., $C M=R$ or $\sqrt{(x-2)^{2}+(y-3)^{2}}=5$, hence $(x-2)^{2}+(y-3)^{2}=25$. We can also write it as: $(x-2)^{2}+(y-3)^{2}-25=0$.
We have obtained the equatio... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,440 |
59. Form the equation of the geometric locus of points that are equally distant from the $O x$ axis and from the point $F(0 ; 2)$. Sketch the curve. | Solution. Let $M(x ; y)$ be a moving point describing the curve (coordinates $x$ and $y$ are current). Then, according to the problem, the distances from point $M$ to the $O x$ axis and to point $F$ are the same (Fig. $10, a$), i.e., $A M=F M$ or
$$
y=\sqrt{x^{2}+(y-2)^{2}}
$$
from which $y^{2}=x^{2}+(y-2)^{2}$.
We ... | \frac{x^{2}}{4}+1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,441 |
73. Find the equation of the line that forms an angle of $135^{\circ}$ with the $O x$ axis and intersects the $O y$ axis at the point $(0 ; 5)$. Determine whether this line passes through the points $A(2 ; 3)$ and $B(2 ;-3)$. Plot the line. | Solution. From the condition of the problem, it follows that the segment cut off by the line on the ordinate axis, $b=5$, the slope $k=\operatorname{tg} 135^{\circ}=-1$. Therefore, by formula (1) we have
$$
y=-x+5
$$
Substituting the coordinates of point $A$ into the desired equation of the line instead of the curren... | -x+5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,442 |
75. Write the equation of a line parallel to the $O x$ axis and cutting off a segment equal to -2 from the $O y$ axis. Check if the points $A(2; -2)$ and $B(3; 2)$ lie on this line. | Solution. According to the problem, the slope $k=0$, the initial ordinate $b=-2$, therefore, the equation of the line we are looking for is $y=-2$. It is easy to see that the coordinates of any point $M(x; -2)$ satisfy this equation.
Point $A$ lies on the line we are looking for, since its coordinates satisfy the equa... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,443 |
76. Write the equation of a line parallel to the $O y$ axis and cutting off a segment of length 3 from the $O \ddot{x}$ axis. Check whether the points $A(3 ; 4)$ and $B(-3 ; 2)$ lie on this line. | Solution. According to the problem, the equation of the line is
$x=3$. Note that the coordinates of any point $M(3 ; y)$ satisfy this equation. Point $A$ lies on the desired line, as its coordinates satisfy the equation of the line $x=3$. Point $B$ does not lie on this line, as its coordinates do not satisfy the equati... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,444 | |
77. Find the equation of the line passing through the point $(2 ; 2)$ and making an angle of $60^{\circ}$ with the $O x$ axis. | Solution. We will use the equation of a straight line passing through a given point in a given direction (4). According to the condition, $k=\operatorname{tg} 60^{\circ}=\sqrt{3}$, therefore, the required equation of the line will be
$$
y--2=\sqrt{3}(x-2) \text { or } y=x \sqrt{ } \overline{3}+2(1-\sqrt{3}) .
$$ | \sqrt{3}x+2(1-\sqrt{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,445 |
78. Find the slope of the line and the segment it intercepts on the y-axis, given that the line passes through the points $M(2, -1)$ and $P(-1, 8)$. | Solution. We will use the equation of a straight line passing through two given points (5). Substituting into the equation instead of $x_{1}, y_{1}, x_{2}, y_{2}$ the coordinates of points $M$ and $P$, we get
$$
\frac{y-(-1)}{8-(-1)}=\frac{x-2}{-1-2}
$$
hence
$$
\frac{y+1}{9}=\frac{x-2}{-3} \text { or } y+1=-3(x-2)
... | k=-3,b=5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,446 |
79. Write the equation of the line that intercepts segments on the coordinate axes, the lengths of which are respectively equal to 3 and -5. | Solution. We will use the equation of a straight line in segments (7). According to the condition, \(a=3, b=-5\), therefore (Fig. 16),
$$
\frac{x}{3}+\frac{y}{-5}=1 \text { or } \frac{x}{3}-\frac{y}{5}=1 .
$$
 A(1 ; 2), B(2 ; 4), C(3 ; 6) ; 2) D(2 ; 3), F(-2 ; 1), F(3 ; 4)$. | Solution. Let's check the fulfillment of condition (8).
$$
\text { 1) } \frac{6-2}{4-2}=\frac{3-1}{2-1} \text { or } \frac{4}{2}=\frac{2}{1} \text {. }
$$
The condition for three given points to lie on the same straight line is satisfied, therefore, points $A, B, C$ lie on the same straight line.
$$
\text { 2) } \fr... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,448 |
81. A straight line intersects the $O x$ axis at some point $M$ and passes through points $A(-2 ; 5)$ and $B(3 ;-3)$. Find the coordinates of point $M$. | Solution. Points $A(-2 ; 5), B(3 ;-3), M(x ; 0)$ lie on the same line, hence condition $(8)$ must be satisfied:
$$
\frac{0-5}{-3-5}=\frac{x-(-2)}{3-(-2)} \text { or } \frac{5}{8}=\frac{x+2}{5} \text {, }
$$
from which $x=1 \frac{1}{8}$. Point $M\left(\frac{9}{8} ; 0\right)$ is the desired one. | (\frac{9}{8};0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,449 |
7. Find the angle between two lines:
1) $y=-2 x, y=3 x+5$
2) $4 x+2 y-5=0,6 x+3 y+1=0$
3) $\sqrt{3} x-y-2=0, \quad \sqrt{3} x+y-1=0$
4) $2 x+3 y-1=0,3 x-2 y+1=0$
5) $3 x+4 y-5=0,5 x-2 y+7=0$
6) $y=2 x+3, y-4=0$
7) $2 x-4 y+7=0, x-5=0$. | Solution.
1) $k_{1}=-2, k_{2}=3$, therefore, by formula (1) we get (Fig. $18, a)$ :
$$
\operatorname{tg} \varphi=\frac{k_{2}-k_{1}}{1+k_{1} k_{2}}=\frac{3-(-2)}{1+(-2) \cdot 3}=-1, \varphi=135^{\circ}
$$
2) $k_{1}=-2, k_{2}=-2$, therefore,
$$
\operatorname{tg} \varphi=\frac{-2-(-2)}{1+(-2) \cdot(-2)}=0, \varphi=0^{... | 6326^{\} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,450 |
110. Write the normal equation of the line if the length of the normal $p=3$, and the angle $\alpha$, formed by it with the positive direction of the $O x$ axis, is $45^{\circ} ; 315^{\circ}$. | Solution. We will use formula (2). According to the condition: 1) $p=3, \alpha=45^{\circ}$, therefore, $x \cdot \cos 45^{\circ}+y \cdot \sin 45^{\circ}-3=0$
or
$$
\frac{\sqrt{2}}{2} x+\frac{\sqrt{2}}{2} y-3=0
$$
2) $p=3, \alpha=315^{\circ}$, therefore,
$$
x \cdot \cos 315^{\circ}+y \cdot \sin 315^{\circ}-3=0
$$
or... | \frac{\sqrt{2}}{2}x+\frac{\sqrt{2}}{2}y-3=0\frac{\sqrt{2}}{2}x-\frac{\sqrt{2}}{2}y-3=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,451 |
111. Reduce the equations of the lines to normal form:
1) $2 x-3 y-10=0$
2) $3 x+4 y=0$ | Solution. In the first equation, $C=-10$, therefore, the normalizing factor $M$ should be taken with a positive sign:
$$
M=\frac{1}{\sqrt{2^{2}+3^{2}}}=\frac{1}{\sqrt{13}} .
$$
Multiplying all terms of the equation by this factor, we obtain the equation of the line in normal form:
$$
\frac{2}{\sqrt{13}} x-\frac{3}{\... | \frac{2}{\sqrt{13}}x-\frac{3}{\sqrt{13}}y-\frac{10}{\sqrt{13}}=0\frac{3}{5}x+\frac{4}{5}0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,452 |
113. Given a triangle with vertices $A(4 ; 6), B(-3 ; 0)$, $C(2 ;-3)$. Find the angles of the triangle, the equation of the bisector $A D$, the equation of the altitude $C E$, and the point of their intersection. | Solution. The angular coefficients of the lines $AB, BC, AC$ (Fig. 19) will be found using formula (6) §5:
$$
\begin{gathered}
k_{A B}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{0-6}{-3-4}=\frac{6}{7} ; k_{B C}=\frac{-3-0}{2-(-3)}=-\frac{3}{5} ; k_{A C}= \\
=\frac{-3-6}{2-4}=\frac{9}{2} .
\end{gathered}
$$
Now we will fin... | F(0;-\frac{2}{3}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,453 |
184. Write the equations of the circles: 1) with center at point $C(-2 ; 3)$ and radius $R=5$; 2) with center at point $C\left(\frac{1}{3} ;-1\right)$ and radius $R=\frac{1}{2}$. | Solution. 1) Given $a=-2, b=3, R=5$, therefore, by formula (3) we obtain the required equation of the circle
$$
(x+2)^{2}+(y-3)^{2}=25
$$
or
$$
x^{2}+4 x+4+y^{2}-6 y+9=25
$$
or
$$
x^{2}+y^{2}+4 x-6 y-12=0
$$
2) Here $a=\frac{1}{3}, b=-1, R=\frac{1}{2}$, therefore, the equation of the circle will be:
or
$$
\begi... | 36x^{2}+36y^{2}-24x+72y+31=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,454 |
185. Reduce the equations of the circles to their standard form:
1) $x^{2}+y^{2}-4 x+6 y-3=0$
2) $4 x^{2}+4 y^{2}-8 x+4 y-11=0$ | Solution. We will bring the first equation to the standard form as follows:
$$
\left(x^{2}-4 x+4\right)+\left(y^{2}+6 y+9\right)-4-9-3=0
$$
from which
$$
(x-2)^{2}+(y+3)^{2}=16
$$
consequently, the center of the circle is at point $C(2 ;-3)$, and the radius $R=4$.
We will bring the second equation to the standard ... | (x-2)^{2}+(y+3)^{2}=16(x-1)^{2}+(y+\frac{1}{2})^{2}=4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,455 |
214. By transforming the coordinates through a parallel shift of the axes, simplify the equation of the circle
$$
x^{2}+y^{2}-4 x+6 y-68=0
$$
taking the new origin as the center of the circle. | Solution. Let's find the center of the circle, i.e., find the origin of the new coordinate system. We will convert the given equation of the circle to its standard form:
$$
(x-2)^{2}+(y+3)^{2}=81
$$
From this, it is clear that the center of the circle lies at the point $C(2, -3)$, i.e., we take the new origin as poin... | X^{2}+Y^{2}=81 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,456 |
219. Determine the coordinates of the focus of the parabola $y=-\frac{1}{6} x^{2}$ and write the equation of its directrix. | Solution. We write the given equation of the parabola in canonical form as $x^{2}=-6 y$. By comparing this equation with equation (7), we get $2 p=-6, p=-3$. Therefore, the focus of the parabola is at the point $F(0; -1.5)$ and the equation of the directrix $DE y=1.5$. | F(0;-1.5),1.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,457 |
221. Find the coordinates of the vertex of the parabola given by the equation $x=2 y^{2}+4 y-5$. Write the equation of the axis of symmetry and plot the parabola. | Solution. Let's find the coordinates of the vertex of the parabola:
$$
x=2\left(y^{2}+2 y-\frac{5}{2}\right)=2\left[(y+1)^{2}-1-\frac{5}{2}\right]=2(y+1)^{2}-7
$$
from which $x+7=2(y+1)^{2}$ or $X=2 Y^{2}$, where $X=x+7$ and $Y=y+1$, i.e., $\alpha=-7, \beta=-1$ (see formulas (4)). Therefore, the vertex of the parabol... | O_{1}(-7;-1),axisofsymmetry:-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,459 |
246. Write the equation of an equilateral hyperbola, given that it passes through the point ( $1 ; 4$ ). Find the coordinates of the vertices of the hyperbola and the distance between them. Sketch the hyperbola. | Solution. According to the problem, $1 \cdot 4 = m$ or $m = 4$, hence the required equation of the hyperbola is $xy = 4$. Since $m = 4 > 0$, the branches of the hyperbola are located in the I and III quadrants. The vertices of the hyperbola are $A(2; 2)$ and $A_{1}(-2; -2)$.
 by a parallel translation of the axes, if the new origin is taken to be the point \( O_{1}(1; 2) \). Find the vertices of the hyperbola, the equations of the real and imaginary axes of symmetry in the new coordinate system, and in the old one. | Solution. We will use formulas (4). Substituting $x=X+1, y=Y+2$ into the given equation, we get:
$$
Y+2=\frac{2 X+2-3}{X+1-1}
$$
from which
$$
Y+2=2+\frac{-1}{X} \text { or } Y=-\frac{1}{X}
$$
In the new coordinate system $X O_{1} Y$, the branches of the hyperbola are located in the II and IV quadrants $(m=-1<0)$. ... | A(0;3),A_{1}(2;1),x+y-3=0,x-y+1=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,461 |
248. The equation of the hyperbola $y=\frac{1-3 x}{2 x-1}$ can be transformed using a parallel translation of the coordinate axes to the form $X Y=m$. Plot this hyperbola. | Solution. In this equation, by dividing the numerator by the denominator, we get:
$$
\begin{gathered}
y=-\frac{3}{2}-\frac{\frac{1}{4}}{x-\frac{1}{2}} \text { or } y+1.5=\frac{-0.25}{x-0.5}, \\
\text { or } Y=-\frac{0.25}{X}, \text { or } X Y=-0.25,
\end{gathered}
$$
where $X=x-0.5$ and $Y=y+1.5$.
The origin of the ... | XY=-0.25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,462 |
284. Given the vertices of a triangle $A(2 ; 0 ; 3), B(-5 ; 2 ; 1), C(3$; $2 ; 1)$. Find the length of its median, drawn from vertex $A$. | Solution. The coordinates of point $D$ (Fig. 34) will be found using formulas (4):
$$
x=\frac{-5+3}{2}=-1 ; y=\frac{2+2}{2}=2 ; \quad z=\frac{1+1}{2}=1
$$
The length of $A D$ will be found using formula (1):
$$
A D=\sqrt{3^{2}+(-2)^{2}+2^{2}}=\sqrt{17} \approx 4.12
$$ | \sqrt{17} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,464 |
303. Vector $\bar{a}=\overline{A B}$, whose length is $a=5$, forms an angle of $60^{\circ}$ with the $O x$ axis (with vector $\overline{O E}=\bar{e}$). Find the projection of this vector on the given axis (Fig. 43).
$ and $B(-3; 2; 1)$. | Solution. $\quad X=x_{2}-x_{1}=-3-1=-4, \quad Y=y_{2}-y_{1}=$ $=2-(-3)=5, Z=z_{2}-z_{1}=1-4=-3$. Therefore, $\overline{A B}=$ $=\{-4 ; 5 ;-3\}$. | {-4;5;-3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,471 |
310. Find the distance between the points $A(3; -4; -1)$ and $B(-1; 2; -3)$. | Solution. Using formula (15), we get
$$
d=|\overline{A B}|=\sqrt{(-1-3)^{2}+(2+4)^{2}+(-3+1)^{2}}=\sqrt{56} \approx 7.48
$$ | 7.48 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,472 |
311. The lengths of vectors $\bar{a}$ and $\bar{b}$ are $a=8, b=5$ and the angle between the vectors $(\hat{a}, \bar{b})=150^{\circ}$. Find the scalar squares of the vectors and their scalar product. | Solution. Using formulas (8) and (10), we obtain the scalar product of the vectors
$$
\bar{a} \cdot \bar{b}=8 \cdot 5 \cos 150^{\circ}=40\left(-\frac{\sqrt{3}}{2}\right)=-20 \sqrt{3}
$$
the scalar squares of the vectors
$$
(\bar{a})^{2}=a^{2}=64,(\bar{b})^{2}=b^{2}=25
$$ | (\bar{})^{2}=64,(\bar{b})^{2}=25,\bar{}\cdot\bar{b}=-20\sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,473 |
312. Find the scalar square of the vector $\bar{a}=2 \bar{i}-\bar{j}-2 \bar{k}$ and its length. | Solution. We will use formulas (10), (13), and (14):
$$
(\bar{a})^{2}=a^{2}=2^{2}+(-1)^{2}+(-2)^{2}=9
$$
from which $a=3$. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,474 |
313. Find the scalar product of the vectors
$$
\bar{p}=\bar{i}-3 \bar{j}+\bar{k}, \bar{q}=\bar{i}+\bar{j}-4 \bar{k}
$$ | Solution. The scalar product of the vectors will be found using formula (12):
$$
\bar{p} \bar{q}=1 \cdot 1+(-3) \cdot 1+1 \cdot (-4)=-6
$$ | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,475 |
314. Find the angle $\varphi$ between two vectors
$$
\bar{a}=\bar{i}+\bar{j}-4 \bar{k} \text { and } \bar{b}=\bar{i}-2 \bar{j}+2 \bar{k} .
$$ | Solution. We will use formula (16):
$$
\cos \varphi=\frac{\bar{a} \bar{b}}{a b}=\frac{1-2-8}{\sqrt{1+1+16} \sqrt{1+4+4}}=-\frac{\sqrt{2}}{2},
$$
therefore, $\varphi=135^{\circ}$. | 135 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,476 |
315. Express through the orths $\bar{i}, \bar{j}, \bar{k}$ the orth $\overline{a^{0}}$ of the vector
$$
\bar{a}=\overline{3 i}-4 \bar{j}+6 \bar{k}
$$ | Solution. It is known that the unit vector of vector $\bar{a}$ is the ratio of this vector to its modulus (see item 4 § 2), i.e.
$$
\bar{a}^{0}=\frac{\bar{a}}{a}=\frac{3 \bar{i}-4 \bar{j}+6 \bar{k}}{\sqrt{9+16+36}}=\frac{3}{7} \bar{i}-\frac{4}{7} \bar{j}+\frac{6}{7} \bar{k}
$$ | \frac{3}{7}\bar{i}-\frac{4}{7}\bar{j}+\frac{6}{7}\bar{k} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,477 |
316. Find the projection of the vector $\bar{a}=3 \bar{i}+4 \bar{j}-\bar{k}$ onto the direction of the vector $\bar{b}=\bar{i}+\bar{j}$. | Solution. It is known that
The cosine of the angle between vectors $\bar{b}$ and $\bar{a}$ is found using formula (16):
$$
\cos (\widehat{b}, \bar{a})=\frac{3 \cdot 1+4 \cdot 1+(-1) \cdot 0... | \frac{7\sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,478 |
317. Find the lengths of the vectors $\bar{a}\{2 ; 5 ;-1\}, \bar{b}\{1 ;-1 ;-3\}$ and the scalar product of these vectors. | Solution. The lengths of the vectors will be found using formula (14):
$a=\sqrt{2^{2}+5^{2}+(-1)^{2}}=\sqrt{30}, b=\sqrt{1^{2}+(-1)^{2}+(-3)^{2}}=\sqrt{11} ;$ the scalar product
$$
\vec{a} \bar{b}=2 \cdot 1+5(-1)+(-1)(-3)=0
$$
Therefore, vectors $\bar{a}$ and $\bar{b}$ are perpendicular. | =\sqrt{30},b=\sqrt{11},\vec{}\cdot\vec{b}=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,479 |
318. Find the angles formed by the vector $\bar{a}\{-2 ; 2 ; 1\}$ with the coordinate axes. | Solution. Using formulas (17), we get: $\cos \alpha=\frac{-2}{\sqrt{(-2)^{2}+2^{2}+1^{2}}}=-\frac{2}{3}, \cos \beta=\frac{2}{3}, \quad \cos \gamma=\frac{1}{3}$, hence $\alpha \approx 131^{\circ} 49^{\prime}, \beta \approx 48^{\circ} 11^{\prime}, \gamma \approx 70^{\circ} 32^{\prime}$. | \alpha\approx13149^{\},\beta\approx4811^{\},\gamma\approx7032^{\} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,480 |
319. The vertices of the triangle are given by the radius vectors $\bar{a}=$ $=\{1 ; 2\}, \bar{b}=\{2 ; 5\}, \bar{c}=\{3 ; 4\}$. Determine the sides and internal angles of the triangle. | Solution. From the condition of the problem, the vertices of the triangle are $A(1 ; 2), B(2 ; 5), C(3 ; 4)$.
Therefore, the vectors coinciding with the sides of the triangle are: $\overline{A B}=\{1 ; 3\}, \overline{A C}=\{2 ; 2\}, \overline{B C}=\{1 ;-1\}$ (Fig. 45). Let's find the lengths of the sides of the triang... | A\approx2554^{\},B\approx646^{\} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,481 |
353. Determine what geometric figures are given by the equations:
1) $x^{2}+y^{2}=R^{2}$
2) $x^{2}+y^{2}+z^{2}=R^{2}$
3) $x y=m$
4) $z-x^{2}=0$. | Solution. 1) The equation defines a right circular cylinder with radius $R$ and axis $O z$. 2) The equation represents a spherical surface with the center coinciding with the origin and radius $R$. 3) The equation represents a cylindrical surface, where the directrix is a hyperbola lying in the plane $x O y$, and the g... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,482 |
354. Check if the points $A\left(0 ; 1 ; \frac{1}{3}\right), B(1 ; 2 ; 4)$, $C(1 ; 1 ; 2)$ lie on the surface given by the equation
$$
y^{3}+2 x y-3 z=0
$$ | Solution. Points $A$ and $B$ lie on the given surface since their coordinates satisfy the equation. Let's verify this. Substituting the values $x=0, y=1, z=\frac{1}{3}$ into the equation, we get $1^{3}+2 \cdot 0 \cdot 1-3 \cdot \frac{1}{3}=0$. If we substitute the values $x=1, y=2, z=4$, we obtain $2^{3}+2 \cdot 1 \cdo... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,483 |
355. Find the equation of the sphere with center at point $C(a ; b ; c)$ and radius $R$. | Solution. By definition, a sphere is the geometric locus of points equally distant from a given point called the center. Therefore, the distance from any point $M(x ; y ; z)$, lying on the sphere, to the center $C(a ; b ; c)$ is $R$, i.e.,
$$
l^{\prime} \overline{(x-a)^{2}+(y-b)^{2}+(z-c)^{2}}=R
$$
hence
$$
(x-a)^{2... | (x-)^{2}+(y-b)^{2}+(z-)^{2}=R^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,484 |
356. Find the center and radius of the sphere
$$
x^{2}+y^{2}+z^{2}-2 x+6 y-6=0
$$ | Solution. We will transform the given equation into the form (1)
\[
\begin{gathered}
x^{2}+y^{2}+z^{2}-2 x+6 y-6=\left(x^{2}-2 x+1\right)+\left(y^{2}+6 y+9\right)+ \\
+z^{2}-1-9-6=(x-1)^{2}+(y+3)^{2}+z^{2}-16=0
\end{gathered}
\]
or
\[
(x-1)^{2}+(y+3)^{2}+z^{2}=16
\]
from which
\[
a=1, b=-3, c=0, R^{2}=16
\]
i.e.,... | C(1,-3,0),R=4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,485 |
367. Determine the geometric figure represented by the system of equations
$$
\left\{\begin{array}{l}
x^{2}+y^{2}+z^{2}-4 x=0 \\
y=1
\end{array}\right.
$$ | Solution. Solving the given system of equations, we obtain the curve $x^{2}+z^{2}-4 x+1=0$, i.e., the circle $(x-2)^{2}+z^{2}=$ $=3$, lying in the plane $y=1$, with radius $R=\sqrt{ } \overline{3}$ and centered at the point $(2 ; 1 ; 0)$. | (x-2)^{2}+z^{2}=3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,486 |
371. Form the equation of the plane $Q$, which intercepts a segment $O A=3$ on the $O x$ axis and is perpendicular to the vector $\bar{n}\{2 ; -3 ; 1\}$. | Solution. According to the problem, point $A$ lies on the plane with coordinates $x=3, y=0, z=0$. Using formula (4), we obtain the equation of the desired plane
$$
2(x-3)-3 y+z=0
$$
or
$$
2 x-3 y+z=6
$$ | 2x-3y+6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,487 |
372. Form the equation of the plane passing through the point $M_{0}(1 ; 2 ; -1)$ and perpendicular to the vector $\bar{n}\{3 ; 0 ; 2\}$. | Solution. Applying formula (4), we get
$$
3(x-1)+0(y-2)+2(z+1)=0
$$
or
$$
3 x+2 z-1=0
$$
The desired equation of the plane does not contain $y$, i.e., it is the equation of a plane parallel to the $O y$ axis. | 3x+2z-1=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,488 |
373. Find the equation of the plane: 1) passing through the axis $O x$ and the point $(1 ; 2 ;-3) ; 2$ ) parallel to the axis $O z$ and passing through the points $(1 ; 0 ; 1)$ and $(-2 ; 1 ; 3)$. | Solution. 1) Each plane passing through the axis $O x$ is defined by an equation of the form
$$
B y+C z=0
$$
We require that the plane passes through the point $(1 ; 2 ;-3):$
$$
2 B-3 C=0 \text { or } B=\frac{3 C}{2} .
$$
Therefore, the equation of the desired plane will be of the form
$$
\frac{3 C}{2} y+C z=0
$$
... | 3y+20 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,489 |
374. Form the equation of the plane passing through the points $M(1 ; 2 ; 0), N(1 ; -1 ; 2), P(0 ; 1 ; -1)$ and find the angles of its normal with the coordinate axes. | Solution. It is known that the position of a plane is determined by three points (not lying on the same line). Let's write the equation of any plane passing through the point $M(1; 2; 0):$
$$
A(x-1)+B(y-2)+C z=0
$$
To obtain the desired equation of the plane, we need to require that the coordinates of points $N$ and ... | 5x-2y-3z-1=0,\alpha\approx3548^{\},\beta\approx10856^{\},\gamma\approx1197^{\} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,490 |
375. Find the distance:
1) from the point $(2 ; 3 ;-4)$ to the plane
$$
2 x+6 y-3 z+16=0
$$
2) from the point $(2 ;-4 ; 1)$ to the plane
$$
x-8 y+4 z=0
$$ | Solution. 1) We will bring the equation to the normal form by multiplying the given equation by the normalizing factor
$$
M=\frac{1}{-\sqrt{4+36+9}}=-\frac{1}{7}
$$
We get
$$
-\frac{2}{7} x-\frac{6}{7} y+\frac{3}{7} z-\frac{16}{7}=0
$$
or
$$
\frac{2 x+6 y-3 z+16}{-7}=0
$$
The distance $d$ will be found using form... | 4\frac{2}{9} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,491 |
376. Determine which of the following pairs of planes intersect, are parallel, or coincide:
1) $4 x-6 y+3 z+5=0 \quad$ and $\quad 2 x-3 y+z-5=0$
2) $6 x+8 y-4 z-6=0$ and $3 x+4 y-2 z+3=0$
3) $3 x-6 y+3 z-6=0$ and $-x+2 y-z+2=0$;
4) $6 x-9 z+5=0 \quad$ and $2 x-3 z+1=0$. | Solution. 1) The condition of parallelism (11) is not satisfied, since the coefficients at the current coordinates are not proportional.
Check:
$$
\frac{A_{1}}{A_{2}}=\frac{4}{2}=2, \frac{B_{1}}{B_{2}}=\frac{-6}{-3}=2, \quad \frac{C_{1}}{C_{2}}=\frac{3}{1}=3
$$
therefore,
$$
\frac{A_{1}}{A_{2}}=\frac{B_{1}}{B_{2}} ... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,492 |
377. Determine that the planes $2 x+3 y-4 z+1=0$ and $5 x-$ $-2 y+z+6=0$ are mutually perpendicular. | Solution. The condition of perpendicularity of planes is satisfied, since
$$
2 \cdot 5 + 3(-2) + (-4) \cdot 1 = 0
$$
Therefore, the planes are mutually perpendicular. 378. Find the angle between the planes:
1) $x + 2y - 2z + 1 = 0$
and $2x + 6y + 3z - 2 = 0$
2) $x - 2y + 3 = 0$
and $y + 2z - 5 = 0$.
Solution. 1) Us... | 6736' | Geometry | proof | Yes | Yes | olympiads | false | 30,493 |
412. Find the direction cosines of the line
$$
\left\{\begin{array}{c}
2 x-3 y-3 z-9=0 \\
x-2 y+z+3=0
\end{array}\right.
$$ | Solution. First, we convert the equations of the line to the canonical form. For this, in the given system of equations, we eliminate the variable $x$ and express $z$ in terms of $y$:
$$
\left\{\begin{array}{c|c}
2 x-3 y-3 z-9=0, & 1 \\
x-2 y+z+3=0 ; & -2
\end{array} \begin{array}{r}
2 x-3 y-3 z-9=0 \\
-2 x+4 y-2 z-6=... | \cos\alpha=\frac{9}{\sqrt{107}},\cos\beta=\frac{5}{\sqrt{107}},\cos\gamma=\frac{1}{\sqrt{107}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,494 |
413. Find the angle between the line
\[
\left\{\begin{array}{l}
x-2 z+3=0 \\
y+3 z-1=0
\end{array}\right.
\]
and the plane
\[
2 x-y+z+3=0
\] | Solution. We will write the equations of the line (A) in canonical form. From the first equation of the system (A) $z=\frac{x+3}{2}$, from the second equation $z=\frac{1-y}{3}$, hence
$$
\frac{x+3}{2}=\frac{1-y}{3}=\frac{z}{1}
$$
or
$$
\frac{x+3}{2}=\frac{y-1}{-3}=\frac{z}{1}
$$
Therefore, the direction coefficient... | \varphi\approx6049^{\} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,495 |
414. The line $L$ passes through the point $(2 ; 1 ;-1)$ and the point of intersection of the line
$$
\frac{x-3}{2}=\frac{y}{-1}=\frac{z+1}{1}
$$
with the plane
$$
x-y+z-1=0
$$
Find the angle formed by the line $L$ with the plane
$$
x+2 y-z+3=0
$$ | Solution. The line (A) and the plane (B) intersect, since the condition of parallelism (14) is not satisfied, i.e., $2 \cdot 1 + (-1) \cdot (-1) + 1 \cdot 1 \neq 0$. Let's find the point of intersection of the line (A) with the plane (B). Write the equations of the line (A) in parametric form:
$$
\frac{x-3}{2}=t, \fra... | 197^{\} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,496 |
450. For what values of $x$ is the equality $|x+6|=$ $=-(x+6)$ true? | Solution. From the definition of absolute value, it follows that $x+6 \leqslant 0$ or $x \leqslant -6$. | x\leqslant-6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,497 |
454. Given the function (rational fractional) $f(x)=\frac{2 x-3}{3 x^{2}-1}$. Find $f(-2) ; f(0) ; f(1)$. | Solution. $f(-2)=\frac{2(-2)-3}{3(-2)^{2}-1}=-\frac{7}{11} ; f(0)=3 ; f(1)=-\frac{1}{2}$. The function $f(x)$ is undefined at points $x_{1}=\frac{1}{\sqrt{3}}$ and $x_{2}=-\frac{1}{\sqrt{3}}$, as the denominator becomes zero at these points. | f(-2)=-\frac{7}{11};f(0)=3;f(1)=-\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,499 |
455. Find the domains of the following functions:
1) $y=\sqrt{3+2 x}$; 2) $f(x)=1+\sqrt{9-x^{2}}$;
2) $\varphi(x)=\sqrt{\lg \frac{5 x-x^{2}}{4}}$; 4) $y=\sqrt{3-x}+\arccos \frac{x-2}{3}$. | Solution. 1) The function is defined at those points where $3+2 x \geqslant 0$ or $2 x \geqslant-3$, from which $x \geqslant-\frac{3}{2}$. Thus, the function is defined on the interval $\left[-\frac{3}{2}, \infty\right)$.
2) The function takes real values at those points where the expression under the square root sign... | [-\frac{3}{2},\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,500 |
456. Are the functions identical:
1) $f(x)=\frac{x}{x^{2}}$ and $g(x)=\frac{1}{x}$
2) $f(x)=x$ and $g(x)=\sqrt{x^{2}}$ ? | Solution. 1) Both functions are defined on the entire number line, except for the point $x=0$. Thus, the domains of definition
of both functions coincide. The functions take the same values fo... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,501 |
482. Prove that the following variable quantities are infinitesimally small:
1) $x_{n}=-\frac{1}{n}$
2) $x_{n}=(-1)^{n-1} \frac{1}{n}$
3) $\beta_{n}=\frac{1}{2 n-1}$ | Solution. Assuming $n=1,2,3,4, \ldots$, let's write the sequence of values of the variables:
1) $-1,-\frac{1}{2},-\frac{1}{3},-\frac{1}{4}, \ldots,-\frac{1}{n}, \ldots$;
2) $1,-\frac{1}{2}, \frac{1}{3},-\frac{1}{4}, \ldots,(-1)^{n-1} \frac{1}{n}, \ldots$;
3) $1, \frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \ldots, \frac{1}{... | proof | Calculus | proof | Yes | Yes | olympiads | false | 30,503 |
483. Using the definition of the limit of a sequence, prove that $a=\frac{2}{3}$ is the limit of the sequence, $x_{n}=$ $=\frac{2 n-1}{3 n+5}$. | Solution. Consider the difference $x_{n}-\frac{2}{3}=\frac{2 n-1}{3 n+5}-\frac{2}{3}=$ $=\frac{-13}{3(3 n+5)}$ and estimate its absolute value
$$
\left|x_{n}-\frac{2}{3}\right|=\left|\frac{-13}{3(3 n+5)}\right|=\frac{13}{3(3 n+5)}
$$
It is easy to see that $\frac{13}{3(3 n+5)}<\frac{13}{3 n+5}<\frac{13}{n}$.
The qua... | proof | Algebra | proof | Yes | Yes | olympiads | false | 30,504 |
484. Prove that $\lim _{x \rightarrow \infty} \frac{2 x-1}{4 x+3}=\frac{1}{2}$.
For what values of $x$ will the values of the function differ from its limit by less than 0.01? | Solution. Consider the absolute value of the difference
$$
|f(x)-A|=\left|\frac{2 x-1}{4 x+3}-\frac{1}{2}\right|=\frac{5}{2|4 x+3|}
$$
In the right-hand side of the equation, the denominator of the fraction is an infinitely large quantity as $x \rightarrow \infty$, while the numerator is a constant. Therefore, the fr... | x>61.75 | Calculus | proof | Yes | Yes | olympiads | false | 30,505 |
497. Find the limits:
1) $\lim _{x \rightarrow 2}\left(3 x^{2}-2 x+7\right)$
2) $\lim _{x \rightarrow 3} \frac{3 x^{2}-1}{2 x^{3}+6 x^{2}-5}$;
3) $\lim _{x \rightarrow 2} \frac{x-2}{x^{2}+3 x-1}$;
4) $\lim _{x \rightarrow 3} \frac{x^{2}-2 x+1}{x-3}$;
5) $\lim _{x \rightarrow 0} \frac{5 x^{3}+2 x^{2}-x}{3 x}$;
6) $\lim ... | Solution. 1) Applying the theorems on limits, we get $\lim _{x \rightarrow 2}\left(3 x^{2}-2 x+7\right)=3 \lim _{x \rightarrow 2} x \cdot \lim _{x \rightarrow 2} x-2 \lim _{x \rightarrow 2} x+\lim _{x \rightarrow 2} 7=3 \cdot 2 \cdot 2-$ $-2 \cdot 2+7=15$
2) The limits of the numerator and denominator exist and the li... | 15,\frac{26}{103},0,\infty,-\frac{1}{3},\frac{2}{3},notfound,0,\frac{2}{5},-1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,506 |
518. Find the limits:
1) $\lim _{x \rightarrow 0} \frac{x}{\sin x}$; 2) $\lim _{x \rightarrow 0} \frac{\sin 4 x}{\sin 5 x}$; 3) $\lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}}$. | Solution. 1) $\lim _{x \rightarrow 0} \frac{x}{\sin x}=\lim _{x \rightarrow 0} \frac{1}{\frac{\sin x}{x}}=\frac{1}{1}=1$.
2) Transform the given expression so that the problem is reduced to limit (1):
$$
\lim _{x \rightarrow 0} \frac{\sin 4 x}{\sin 5 x}=\lim _{x \rightarrow 0} \frac{\frac{\sin 4 x}{4 x} \cdot 4 x}{\f... | 1,\frac{4}{5},\frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,507 |
2. Find the limits:
1) $\lim _{x \rightarrow 3}\left(\frac{1}{x-3}-\frac{6}{x^{2}-9}\right)$
2) $\lim _{x \rightarrow \infty}\left(\sqrt{x^{2}+1}-x\right)$
3) $\lim _{n \rightarrow \infty} 2^{\prime \prime} \sin \frac{x}{2^{n}}$
4) $\lim _{x \rightarrow 1}(1-x) \operatorname{tg} \frac{\pi x}{2}$. | Solution. 1) It is not hard to see that in this example we have the difference of two infinitely large positive quantities, i.e., an indeterminate form of the type $\infty-\infty$:
$\lim _{x \rightarrow 3}\left(\frac{1}{x-3}-\frac{6}{x^{2}-9}\right)=\lim _{x \rightarrow 3} \frac{x-3}{x^{2}-9}=\lim _{x \rightarrow 3} \... | \frac{1}{6},0,x,\frac{2}{\pi} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,508 |
552. Find the limits:
1) $\lim _{x \rightarrow \infty}\left(1+\frac{a}{x}\right)^{x}$
2) $\lim _{n \rightarrow \infty}\left(\frac{n-1}{n+2}\right)^{2 n+1}$
3) $\lim _{x \rightarrow 0} \sqrt[x]{1+3 x}$
4) $\lim _{x \rightarrow 0}(\cos 2 x)^{\frac{1}{\sin ^{2} x}}$ | Solution. We have an indeterminate form of $1^{\infty}$. Each of the given expressions under the limit sign will be transformed so that the problem reduces to the second remarkable limit.
1) $\lim _{x \rightarrow \infty}\left(1+\frac{a}{x}\right)^{x}=\lim _{x \rightarrow \infty}\left[\left(1+\frac{a}{x}\right)^{\frac{... | e^,\frac{1}{e^6},e^3,\frac{1}{e^2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,509 |
582. Using Theorem 5 of this paragraph, find the limits:
1) $\lim _{x \rightarrow 0} \frac{\sin 3 x}{5 x}$;
2) $\lim _{x \rightarrow 0} \frac{\sin 9 x}{\sin 5 x}$
3) $\lim _{x \rightarrow 0} \frac{\cos x-\cos 2 x}{x^{2}}$;
4) $\lim _{x \rightarrow 0} \frac{\tan p x}{\tan q x}$; 5) $\lim _{x \rightarrow 0} \frac{\ln (1+... | Solution. 1) $\sin 3 x \sim 3 x$, therefore, $\lim _{x \rightarrow 0} \frac{\sin 3 x}{5 x}=$ $=\lim _{x \rightarrow 0} \frac{3 x}{5 x}=\frac{3}{5}$. The infinitesimal quantities $\beta=\sin 3 x$ and $\alpha=5 x$ are of the same order.
2) $\lim _{x \rightarrow 0} \frac{\sin 9 x}{\sin 5 x}=\lim _{x \rightarrow 0} \frac{... | \frac{3}{5},\frac{9}{5},\frac{3}{2},\frac{p}{q},\frac{3}{7},\frac{5}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,511 |
585. Find the one-sided limits of the function
$$
f(x)=\operatorname{arctg} \frac{1}{x-1}(x \neq 1) \text { as } x \rightarrow 1
$$ | Solution. $f(1+0)=\lim _{\substack{x \rightarrow 1+0 \\(x>1)}} \operatorname{arctg} \frac{1}{x-1}=\lim _{z \rightarrow+\infty} \operatorname{arctg} z=\frac{\pi}{2}$, where $z=\frac{1}{x-1}$; from this it is clear that if $x \rightarrow 1+0$, then $z \rightarrow+\infty$.
$$
f(1-0)=\lim _{\substack{x \rightarrow 1-0 \\(... | f(1+0)=\frac{\pi}{2},\quadf(1-0)=-\frac{\pi}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,512 |
595. Using the definition of continuity of a function, prove that: 1) the function $f(x)=x^{2}+3$ is continuous at the point $x=2$; 2) the function $f(x)=\frac{1}{x-5}(x \neq 5)$ is continuous at the point $x=3$. | Solution. 1) We will use the definition in point 1: a function $f(x)$ is continuous at a point $x$ if an infinitely small increment of the argument $\Delta x$ corresponds to an infinitely small increment of the function $\Delta y$, i.e., if $\Delta x \rightarrow 0$, then $\Delta y \rightarrow 0$ (see Fig. 56).
Let's f... | proof | Calculus | proof | Yes | Yes | olympiads | false | 30,513 |
597. Using the properties of continuous functions, prove the continuity for any $x$ of the following functions:
$$
\text { 1) } f(x)=x^{4}+3 x+5 ; \quad 2) \quad f(x)=x^{2} \sin x-\frac{x^{2}}{x^{2}+3}
$$ | Solution. 1) The function is continuous at each point $x$, as it consists of the sum of continuous functions (see point 2). The function $x^{4}$ is continuous as the product of continuous functions $x x x x=x^{4}$. The function $3 x$ is also continuous as the product of two continuous functions. The third term 5 is a c... | proof | Calculus | proof | Yes | Yes | olympiads | false | 30,515 |
599. Investigate the continuity of the functions:
1) $y=\frac{1}{x^{2}+4}$
2) $y=\frac{1}{x^{2}-4}$
3) $y=\frac{x+3}{x-1}$
4) $y=\cos 3 x$
5) $y=\frac{x^{2}-9}{x^{2}-2 x+4}$
6) $y=\left\{\begin{array}{l}\sin \frac{1}{x} \text { for } x \neq 0, \\ 0 \quad \text { for } x=0 .\end{array}\right.$ | Solution. 1) The function is elementary, defined and continuous on the interval $(-\infty,+\infty)$.
2) The function is algebraic, fractional, defined on the set of all real numbers, except for the points where the denominator of the fraction is zero: $x^{2}-4=0$, hence $x_{1}=2, x_{2}=-2$. At these points, the functi... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,517 |
607. A point moves along a straight line according to the law $s=3 t^{2}+2 t$, where $t$ - time in seconds, $s$ - distance in meters. Find the velocity of the point at the moments $t=3$ and $t=4$. | Solution. Guided by point 3, we sequentially find:
1) $s+\Delta s=3(t+\Delta t)^{2}+2(t+\Delta t)$
2) $\Delta s=3(t+\Delta t)^{2}+2(t+\Delta t)-\left(3 t^{2}+2 t\right)=6 t \Delta t+2 \Delta t+3 \Delta t^{2}$;
3) $v_{\mathrm{cp}}=\frac{\Delta s}{\Delta t}=\frac{\Delta t(6 t+2+3 \Delta t)}{\Delta t}=6 t+2+3 \Delta t$
... | v=20(/)at=3,v=26(/)at=4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,519 |
608. Using the definition of the derivative, find the derivatives of the following functions:
1) $y=3 x+5$
2) $f(x)=\frac{1}{x}$ at the point $x=2$;
3) $f(x)=\sin (2 x-3)$ at the point $x=1$. | Solution. 1) Give the argument $x$ an increment $\Delta x$, then we get a new value of the function
$$
y+\Delta y=3(x+\Delta x)+5
$$
Subtract the initial value of the function 1 from this equation. We get the increment of the function $\Delta y=3 \Delta x$. Now find the limit of the ratio $\frac{\Delta y}{\Delta x}$ ... | 3,-\frac{1}{4},2\cos1,4,0,6 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,520 |
623. Find $f^{\prime}(0), f^{\prime}(1), f^{\prime}(2)$, if $f(x)=2 x^{3}+4 x-5$. | Solution. First, we find the derivative in general form. By rule II, and then I, we will have
$$
f^{\prime}(x)=\left(2 x^{3}\right)^{\prime}+(4 x)^{\prime}-(5)^{\prime}=2\left(x^{3}\right)^{\prime}+4(x)^{\prime}-(5)^{\prime}
$$
Using the derivative formulas (1), (2), (3), we finally get
$$
f^{\prime}(x)=6 x^{2}+4
$$... | f^{\}(0)=4;\quadf^{\}(1)=10;\quadf^{\}(2)=28 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,521 |
624. Using formulas and rules of differentiation, find the derivatives of the following functions:
1) $y=x^{\frac{5}{6}}+7$
2) $y=\frac{3 x^{3}}{\sqrt[5]{x^{2}}}-\frac{9}{\sqrt[3]{x^{2}}}+2 \sqrt[6]{x^{5}}$
3) $y=x \sqrt[3]{x}+\frac{1}{\sqrt{x}}+0.1 x^{10}$
4) $y=\left(2 x^{3}+\sqrt{5}\right) 7^{x}$
5) $y=\frac{x}{2-\c... | Solution. 1) $y^{\prime}=\left(x^{\frac{5}{6}}\right)^{\prime}+(7)^{\prime}=\frac{5}{6} x^{-\frac{1}{6}} \quad$ [II; (3), (1)].
2) $y^{\prime}=\left(3 x^{\frac{13}{5}}\right)^{\prime}-\left(9 x^{-\frac{2}{3}}\right)^{\prime}+\left(2 x^{\frac{5}{6}}\right)^{\prime}=\frac{39}{5} x^{\frac{8}{5}}+\frac{18}{3} x^{-\frac{5}{... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,522 |
625. Find the derivative of the function $y=x \tan x+\cot x$. | Solution. $y^{\prime}=(x \tan x)^{\prime}+(\cot x)^{\prime}=(x)^{\prime} \tan x+x(\tan x)^{\prime}+$ $+(\cot x)^{\prime}=\tan x+\frac{x}{\cos ^{2} x}-\frac{1}{\sin ^{2} x}$ [II, III, ; (2), (8), (9)] | \tanx+\frac{x}{\cos^2x}-\frac{1}{\sin^2x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,523 |
626. Find $f^{\prime}(0)$, if $f(x)=e^{x} \arcsin x+\operatorname{arctg} x$. | Solution. $\quad f^{\prime}(x)=\left(e^{x} \arcsin x\right)^{\prime}+(\operatorname{arctg} x)^{\prime}=\left(e^{x}\right)^{\prime} \arcsin x+$ $+e^{x}(\arcsin x)^{\prime}+(\operatorname{arctg} x)^{\prime}=e^{x} \arcsin x+\frac{e^{x}}{\sqrt{1-\lambda^{2}}}+$ $+\frac{1}{x^{2}+1}=e^{x} \frac{1+\sqrt{1-x^{2}} \arcsin x}{\s... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,524 |
627. Calculate the derivative of the function $y=x^{3} \log _{5} x$. | Solution. $y^{\prime}=\left(x^{3}\right)^{\prime} \log _{5} x+x^{3} \cdot\left(\log _{5} x\right)^{\prime}=3 x^{2} \log _{5} x+$ $+x^{3} \cdot \frac{1}{x \ln 5}=x^{2}\left(3 \log _{5} x+\frac{1}{\ln 5}\right)$.
Using formulas and rules of differentiation, find the derivatives of the given functions.
a) Algebraic func... | x^{2}(3\log_{5}x+\frac{1}{\ln5}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,525 |
677. $y=\left(\frac{x+1}{x-1}\right)^{2}$, i.e. $y=u^{2}$, where $u=\frac{x+1}{x-1}$. | Solution. $y^{\prime}=2$ or $^{\prime} y^{\prime}=2 \frac{x+1}{x-1}\left(\frac{x+1}{x-1}\right)^{\prime}=2 \frac{x+1}{x-1} \times$ $\times \frac{(x+1)^{\prime}(x-1)-(x+1)(x-1)^{\prime}}{(x-1)^{2}}=2 \frac{x+1}{x-1} \cdot \frac{(x-1)-(x+1)}{(x-1)^{2}}=$ $=-\frac{4(x+1)}{(x-1)^{3}}[(3) ; I V, I I ;(1),(2) \S 2]$. | -\frac{4(x+1)}{(x-1)^{3}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,526 |
678. $y=(2-5 x)^{100}$, i.e. $y=u^{100}$, where $u=2-5 x$. | Solution. $y^{\prime}=100 u^{99} u^{\prime}$ or $y^{\prime}=100(2-5 x)^{94}(2-5 x)^{\prime}=$ $=-500(2-5 x)^{99}$ I(3); II, I; (1), (2) § 2]. | -500(2-5x)^{99} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,527 |
679. $y=10^{2 x-3}$, i.e. $y=10^{u}$, where $u=2 x-3$. | Solution. $y^{\prime}=10^{\prime \prime} \ln 10 u^{\prime}$ or $y^{\prime}=10^{2 x-3} \ln 10(2 x-3)^{\prime}=$ $=2 \cdot 10^{2 x-3} \ln 10[(4) ; 11,1 ;(1),(2) \S 2]$. | 2\cdot10^{2x-3}\ln10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,528 |
680. $y=\log _{5}\left(x^{3}-1\right)$, i.e. $y=\log _{5} u$, where $u=x^{3}-1$. | Solution. $y^{\prime}=\frac{u^{\prime}}{u \ln 5}$ or $y^{\prime}=\frac{\left(x^{3}-1\right)^{\prime}}{\left(x^{3}-1\right) \ln 5}=\frac{3 x^{2}}{\left(x^{3}-1\right) \ln 5}$ [(5); II]. 681. $y=\cos 5^{x}$, i.e., $y=\cos u$, where $u=5^{x}$.
Solution. $y^{\prime}=-\sin u \cdot u^{\prime}$ or $y^{\prime}=-\sin 5^{x}\lef... | \frac{3x^{2}}{(x^{3}-1)\ln5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,529 |
682. $y=\tan 5 x$, i.e., $y=\tan u$, where $u=5 x$. | Solution. $y^{\prime}=\frac{u^{\prime}}{\cos ^{2} u}$ or $y^{\prime}=\frac{(5 x)^{\prime}}{\cos ^{2} 5 x}=\frac{5}{\cos ^{2} 5 x}=5 \sec ^{2} 5 x$. 683. $y=\operatorname{arctg}^{\prime} \bar{x}$, i.e., $y=\operatorname{arctg} u$, where $u=\sqrt{x}$.
Solution. $y^{\prime}=\frac{u^{\prime}}{1+u^{2}}$ or $y^{\prime}=\fra... | 5\^{2}5x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,530 |
684. $y=\arccos \frac{2 x-1}{\sqrt{3}}$, i.e., $y=\arccos u$, where $u=\frac{2 x-1}{\sqrt{3}}$. | Solution. $y^{\prime}=-\frac{u^{\prime}}{\sqrt{1-u^{2}}}$ or $y^{\prime}=$
$$
=-\frac{1}{\sqrt{1-\left(\frac{2 x-1}{\sqrt{3}}\right)^{2}}}\left(\frac{2 x-1}{\sqrt{3}}\right)^{\prime}=\frac{-\sqrt{3}}{\sqrt{3-(2 x-1)^{2}}} \cdot \frac{2}{\sqrt{3}}=
$$
$$
=-\frac{2}{\sqrt{3-(2 x-1)^{2}}}[(11) ; \mathrm{I}, \mathrm{I}]
... | -\frac{2}{\sqrt{3-(2x-1)^{2}}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,531 |
685. $y=\sin ^{2}(2 x-1)$, i.e. $y=u^{2}$, where $u=\sin (2 x-1)$. | Solution. $y^{\prime}=2 u u^{\prime}$ or $y^{\prime}=2 \sin (2 x-1) \cdot[\sin (2 x-1)]^{\prime}=$ $=2 \sin (2 x-1) \cdot \cos (2 x-1)(2 x-1)^{\prime}=4 \sin (2 x-1) \cdot \cos (2 x-1)=$ $=2 \sin 2(2 x-1)$. | 2\sin2(2x-1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,532 |
686. $y=\ln ^{3}(5 x+2)$, i.e. $y=u^{3}$, where $u=\ln (5 x+2)$. | Solution. $y^{\prime}=3 u^{2} u^{\prime}$ or $y^{\prime}=3 \ln ^{2}(5 x+2)[\ln (5 x+2)]^{\prime}=$ $=3 \ln ^{2}(5 x+2) \frac{(5 x+2)^{\prime}}{5 x+2}=\frac{15 \ln ^{2}(5 x+2)}{5 x+2}$. | \frac{15\ln^{2}(5x+2)}{5x+2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,533 |
687. $f(x)=\ln \sqrt{\frac{1-\sin \alpha}{1+\sin \alpha}} ;$ calculate $f^{\prime}\left(\frac{\pi}{4}\right)$. | Solution. Here it is appropriate to first take the logarithm. After logarithmization, we get
$$
f(x)=\frac{1}{2}[\ln (1-\sin \alpha)-\ln (1+\sin \alpha)]
$$
Now we compute the derivative
$$
f^{\prime}(x)=\frac{1}{2}\left[\frac{-\cos \alpha}{1-\sin \alpha}-\frac{\cos \alpha}{1+\sin \alpha}\right]=-\frac{1}{\cos \alph... | -\sqrt{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,534 |
688. $r=\ln \frac{e^{t}}{e^{t}+1} ;$ calculate $r^{\prime}(0)$. | Solution. $r=\ln e^{t}-\ln \left(e^{t}+1\right)=t-\ln \left(e^{t}+1\right)$;
$$
\frac{d r}{d t}=1-\frac{\left(e^{t}+1\right)^{\prime}}{e^{t}+1}=1-\frac{e^{t}}{e^{t}+1}=\frac{1}{e^{t}+1}
$$
Setting $t=0$, we find $r^{\prime}(0)=\frac{1}{e^{0}+1}=\frac{1}{2}$. | \frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,535 |
689. $y=\sin ^{2}(a x+b)$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
689. $y=\sin ^{2}(a x+b)$. | Solution. $y^{\prime}=2 \sin (a x+b)[\sin (a x+b)]^{\prime}=$
$=2 \sin (a x+b) \cos (a x+b)(a x+b)^{\prime}=$
$=2 \sin (a x+b) \cos (a x+b) a=a \sin 2(a x+b)$. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,536 | |
732. Find the derivative of the exponential-power function
$$
y=u^{v}(u>0) \text {, where } u=f(x), v=\varphi(x) \text {. }
$$ | Solution. First, we take the logarithm of the given function (with base e): $\ln y=v \ln u$. Now we differentiate both sides of the equation using the rule for differentiating a composite function. We get
$$
\frac{y^{\prime}}{y}=v^{\prime} \ln u+v \frac{u^{\prime}}{u}
$$
from which
$$
y^{\prime}=y\left(v^{\prime} \l... | y^{\}=u^{v}(v^{\}\lnu+\frac{v}{u}u^{\}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,537 |
733. Find $y^{\prime}$, if $y=x^{x}$. | Solution. $\ln y=x \ln x$.
The derivative of the left side of the equation is found using the rule for differentiating a composite function, and the derivative of the right side is found using the rule for differentiating a product:
$$
\frac{y^{\prime}}{y}=x^{\prime} \ln x+x(\ln x)^{\prime}
$$
or
$$
\frac{y^{\prime... | y^{\}=x^{x}(1+\lnx) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,538 |
734. Find $y^{\prime}$, if
$$
y=(x-1) \sqrt[3]{(5 x+1)^{2}(x+1)}
$$ | Solution. $\ln y=\ln (x-1)+\frac{2}{3} \ln (5 x+1)+\frac{1}{3} \ln (x+1)$;
$$
\frac{y^{\prime}}{y}=\frac{1}{x-1}+\frac{10}{3(5 x+1)}+\frac{1}{3(x+1)}=\frac{2\left(15 x^{2}+7 x-4\right)}{3\left(x^{2}-1\right)(5 x+1)}
$$
from here
$$
\begin{gathered}
y^{\prime}=\frac{2\left(15 x^{2}+7 x-4\right)}{3\left(x^{2}-1\right)... | \frac{2(15x^{2}+7x-4)}{3\sqrt[3]{(x+1)^{2}(5x+1)}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,539 |
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